ISAT Grade 8 Expressions and Equations. Practice it free.

$4 \times 4 \times 4 = 64$, so $\sqrt[3]{64} = 4$.

The Product of Powers Property says that powers with the same base are multiplied by adding the exponents, written $x^{m} \cdot x^{n} = x^{m+n}\textsf{.}$ The numerator is $2^{7} \cdot 2^{4} = 2^{7+4} = 2^{11}\textsf{.}$ The Quotient of Powers Property then subtracts the exponents of powers with the same base, written $\dfrac{x^{m}}{x^{n}} = x^{m-n}\textsf{.}$ So $\dfrac{2^{11}}{2^{6}} = 2^{11-6} = 2^{5}$ and $n = 5\textsf{.}$

Substitute $t = 2$: $f(2) = 8e^{6} + 50$. Since $e^{6} \approx 403.4$, $f(2) \approx 8(403.4) + 50 = 3{,}277 \approx 3 \times 10^{3}$.

Moving the decimal in $0.0075$ three places right gives $7.5$, so $c = 7.5$.

Multiply coefficients and add exponents: $4 \cdot 5 = 20$ and $3+4=7$, giving $20 \times 10^{7}$. Rewrite as $2 \times 10^{1} \times 10^{7}=2 \times 10^{8}\textsf{.}$

The first equation gives $y = -2x + 5\textsf{.}$ The second simplifies to $y = -2x + 6\textsf{,}$ so the slopes match but the intercepts differ and the system has no solution.

Rewrite in slope-intercept form: $y = -\dfrac{3}{4}x + \dfrac{5}{2}$ and $y = \dfrac{1}{2}x - \dfrac{1}{2}$. The slopes $-\dfrac{3}{4}$ and $\dfrac{1}{2}$ are different, so the lines intersect at exactly one point.

Distribute $\dfrac{1}{2}$: $2x + 3 + x = 15$. Combine like terms: $3x + 3 = 15$. Subtract $3$ from both sides: $3x = 12$. Divide by $3$: $x = 4\textsf{.}$